x+y+z=1
4x-4y-3z=0
3x-7y+3z=0
x+3y-6z=0
4x-4y-3z=0
3x-7y+3z=0
x+3y-6z=0
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Use the equations in pairs to eliminate one variable first.
Multiply through equation 1 by 3 to get:
3x + 3y + 3z = 3
Add that to equation 2:
3x + 3y + 3z = 3
+(4x - 4y - 3z = 0)
-----------------------
7x - y = 3
Multiply through equation 3 by 2 to get:
6x - 14y + 6z = 0
Add that to equation 4 to eliminate z again:
6x - 14y + 6z = 0
+(x + 3y - 6z = 0)
---------------------
7x - 11y = 0
Use those two equations to eliminate another variable:
7x - y = 3
7x - 11y = 0
Since the x coefficients are the same, just subtract the second one from the first:
7x - y = 3
-(7x - 11y = 0)
--------------------
10y = 3
Divide both sides by 10:
y = 3/10
Substitute 3/10 for y in the original equations and pick a pair of them to use to eliminate x or z and solve for the other.
Multiply through equation 1 by 3 to get:
3x + 3y + 3z = 3
Add that to equation 2:
3x + 3y + 3z = 3
+(4x - 4y - 3z = 0)
-----------------------
7x - y = 3
Multiply through equation 3 by 2 to get:
6x - 14y + 6z = 0
Add that to equation 4 to eliminate z again:
6x - 14y + 6z = 0
+(x + 3y - 6z = 0)
---------------------
7x - 11y = 0
Use those two equations to eliminate another variable:
7x - y = 3
7x - 11y = 0
Since the x coefficients are the same, just subtract the second one from the first:
7x - y = 3
-(7x - 11y = 0)
--------------------
10y = 3
Divide both sides by 10:
y = 3/10
Substitute 3/10 for y in the original equations and pick a pair of them to use to eliminate x or z and solve for the other.
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It's unusual to have three variables in four equations given...
x+y+z=1 -- (1)
4x-4y-3z=0 -- (2)
3x-7y+3z=0 -- (3)
x+3y-6z=0 -- (4)
(2)+(3) => 7x - 11y = 0
=> 7x = 11y
=> y = 7/11 x
(1)*4 + 2 => 8x + z = 4
z = 4 - 8x
plug in to (4)
x + 21/11 x + 48x - 24= 0
32/11 x + 48x = 24
24 = (528 + 32)/11 x
24 = 560/11 x
x = 24*11/560 = 33/70
y = 7/11 x = 3/10
z = 4 - 8*33/70 = 8/35
===========================
Strange, I checked my answer, it seems to be in correct. Neither "Ovidiu"'s answer is correct.
As I said, this question itself may not be correct in the beginning....
===========================
I took another three equations, #1, #3 and #4, to solve for x, y and z.
I got x = 277/478, y = 153/478 and z = 24/239.
plug in back to #2, the answer is wrong again.
I think Eric should check the question again!!
===========================
x+y+z=1 -- (1)
4x-4y-3z=0 -- (2)
3x-7y+3z=0 -- (3)
x+3y-6z=0 -- (4)
(2)+(3) => 7x - 11y = 0
=> 7x = 11y
=> y = 7/11 x
(1)*4 + 2 => 8x + z = 4
z = 4 - 8x
plug in to (4)
x + 21/11 x + 48x - 24= 0
32/11 x + 48x = 24
24 = (528 + 32)/11 x
24 = 560/11 x
x = 24*11/560 = 33/70
y = 7/11 x = 3/10
z = 4 - 8*33/70 = 8/35
===========================
Strange, I checked my answer, it seems to be in correct. Neither "Ovidiu"'s answer is correct.
As I said, this question itself may not be correct in the beginning....
===========================
I took another three equations, #1, #3 and #4, to solve for x, y and z.
I got x = 277/478, y = 153/478 and z = 24/239.
plug in back to #2, the answer is wrong again.
I think Eric should check the question again!!
===========================
-
x=6z-3y from the last equation
6z-3y+y+z=1
4(6z-3y)-4y-3z=0
7z-2y=0
24z-12y-4y-3z=0
7z-2y=0
21z-16y=0
-21z+6y=0
21z-16y=0
-10y=0 =>y=0
x+z=1
-x+6z=0
7z=1
z=1/7
x=6/7
proof
6/7+3*0-6(1/7)=6/7-6/7=0
x=6/7
y=0
z=1/7
x-6z=0
6z-3y+y+z=1
4(6z-3y)-4y-3z=0
7z-2y=0
24z-12y-4y-3z=0
7z-2y=0
21z-16y=0
-21z+6y=0
21z-16y=0
-10y=0 =>y=0
x+z=1
-x+6z=0
7z=1
z=1/7
x=6/7
proof
6/7+3*0-6(1/7)=6/7-6/7=0
x=6/7
y=0
z=1/7
x-6z=0